Show that $X^3+X^2+1$ has only one real root Consider the polynomial $X^3+X^2+1 \in \mathbb R [X]$. Since it is of odd degree, it has at least one real root. How can I show that it's the only one?
 A: There is no positive root as all terms are positive. Zero is not a root. To check for negative roots, consider $-x^3+x^2+1$, which has only one sign change, so by Descartes rule of signs, the polynomial has exactly one negative root.
A: Derivative being equal to $0$
$$3x^2+2x=0$$
gives $x_1=0$ and $x_2=\frac{-2}{3}$ and in our initial polynomial we have
$p(0)=1$ and $p(\frac{-2}{3})>0$ hence it never crosses the $0$ line again and cannot have any additional zeros amongst the real
A: There are at least two possible approaches:


*

*Let $f(x)=x^3+x^2+1$. Differentiate $f$ to find the maxima and minima; $f$ will have three real roots iff the local maximum is above the $x$-axis and the local minimum is below the axis.

*If you want an algebraic proof, suppose for the sake of contradiction that the polynomial has real roots $a,b,c$. Then $abc=-1$ by Vieta's formulas, so at least one of $a,b,c$ has absolute value at least 1. But 
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=1,$$ again by Vieta's, and this is a contradiction since none of $a,b,c$ is $0$.
